检查两个3Dnumpy数组是否包含重叠的2D数组

我们可以利用views我在一些问答中使用过的辅助功能来发挥作用。要获得子数组的存在，我们可以np.isin在视图上使用或使用更加费力的视图np.searchsorted。

使用np.isin-

# https://stackoverflow.com/a/45313353/ @Divakar
def view1D(a, b): # a, b are arrays
a = np.ascontiguousarray(a)
b = np.ascontiguousarray(b)
void_dt = np.dtype((np.void, a.dtype.itemsize * a.shape[1]))
return a.view(void_dt).ravel(), b.view(void_dt).ravel()
def isin_nd(a,b):
# a,b are the 3D input arrays to give us "isin-like" functionality across them
A,B = view1D(a.reshape(a.shape[0],-1),b.reshape(b.shape[0],-1))
return np.isin(A,B)


我们也可以利用np.searchsorted于views-

def isin_nd_searchsorted(a,b):
# a,b are the 3D input arrays
A,B = view1D(a.reshape(a.shape[0],-1),b.reshape(b.shape[0],-1))
sidx = A.argsort()
sorted_index = np.searchsorted(A,B,sorter=sidx)
sorted_index[sorted_index==len(A)] = len(A)-1
idx = sidx[sorted_index]
return A[idx] == B


因此，这两个解决方案为我们提供了ain中每个子数组的存在掩码b。因此，为了获得所需的计数，它应该是-isin_nd(a,b).sum()或isin_nd_searchsorted(a,b).sum()。

样品运行-

In [71]: # Setup with 3 common "subarrays"
...: np.random.seed(0)
...: a = np.random.randint(0,9,(10,4,5))
...: b = np.random.randint(0,9,(7,4,5))
...:
...: b[1] = a[4]
...: b[3] = a[2]
...: b[6] = a[0]
In [72]: isin_nd(a,b).sum()
Out[72]: 3
In [73]: isin_nd_searchsorted(a,b).sum()
Out[73]: 3


大型阵列上的时间-

In [74]: # Setup
...: np.random.seed(0)
...: a = np.random.randint(0,9,(100,100,100))
...: b = np.random.randint(0,9,(100,100,100))
...: idxa = np.random.choice(range(len(a)), len(a)//2, replace=False)
...: idxb = np.random.choice(range(len(b)), len(b)//2, replace=False)
...: a[idxa] = b[idxb]
# Verify output
In [82]: np.allclose(isin_nd(a,b),isin_nd_searchsorted(a,b))
Out[82]: True
In [75]: %timeit isin_nd(a,b).sum()
10 loops, best of 3: 31.2 ms per loop
In [76]: %timeit isin_nd_searchsorted(a,b).sum()
100 loops, best of 3: 1.98 ms per loop